Inclined & Horizontal Plate Natural Convection
Analyze natural convection boundary layers on flat plates oriented at arbitrary angles from vertical.
Effective Gravity and Angle
For inclined plates ($\theta < 90^\circ$ from vertical), natural convection correlations are modified by replacing the standard gravity constant $g$ with the effective parallel component: $$g_{eff} = g \cos \theta$$ For horizontal plates ($\theta = 90^\circ$), fluid buoyancy forces form rising plumes (on upper hot/lower cold surfaces) or stable stratification layers (on lower hot/upper cold surfaces).
Parameters Setup
Results & Curves
📈 Nusselt number vs Temperature difference (dT)
Engine Output
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INCLINED & HORIZONTAL PLATE CONVECTION
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Rayleigh Number Ra = 4.1974E+08
Nusselt Number Nu = 94.1309
Coeff h = 4.9513 W/m2K
Transfer Q = 40.8481 W
--- DELTA-T SWEEP ---
dT[C] Ra Nu h[W/m2K] Q[W]
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1.00 7.632E+06 28.913 1.5208 0.2281
7.83 5.978E+07 52.419 2.7572 3.2397
14.67 1.119E+08 63.154 3.3219 7.3082
21.50 1.641E+08 70.825 3.7254 12.0143
28.33 2.162E+08 76.966 4.0484 17.2058
35.17 2.684E+08 82.165 4.3219 22.7978
42.00 3.205E+08 86.713 4.5611 28.7349
48.83 3.727E+08 90.781 4.7751 34.9774
55.67 4.248E+08 94.477 4.9695 41.4955
62.50 4.770E+08 97.877 5.1483 48.2655
69.33 5.291E+08 101.032 5.3143 55.2684
76.17 5.813E+08 103.982 5.4694 62.4883
83.00 6.334E+08 106.757 5.6154 69.9117
89.83 6.856E+08 109.380 5.7534 77.5270
96.67 7.377E+08 111.871 5.8844 85.3244
103.50 7.899E+08 114.246 6.0093 93.2948
110.33 8.420E+08 116.516 6.1287 101.4304
117.17 8.942E+08 118.692 6.2432 109.7243
124.00 9.463E+08 120.784 6.3532 118.1701
130.83 9.985E+08 122.799 6.4592 126.7621
137.67 1.051E+09 124.743 6.5615 135.4950
144.50 1.103E+09 126.624 6.6604 144.3642
151.33 1.155E+09 128.444 6.7562 153.3650
158.17 1.207E+09 130.210 6.8490 162.4936
165.00 1.259E+09 131.925 6.9392 171.7462
Calculation Methodology
Mathematical Model & Theory
Natural convection on inclined flat plates is calculated using Churchill & Chu's vertical plate correlation with modified parallel gravity: $$g_{eff} = g \cos \theta$$ $$\text{Nu}_L = \left\{0.825 + \frac{0.387 \text{Ra}_L^{1/6}}{\left[1 + (0.492/\text{Pr})^{9/16}\right]^{8/27}}\right\}^2$$ For horizontal plates ($\theta = 90^\circ$), characteristic length is defined as: $$L_c = \frac{\text{Area}}{\text{Perimeter}} = \frac{L \cdot W}{2(L+W)}$$ Correlations for upper surface of hot plate / lower surface of cold plate: $$\text{Nu} = 0.54 \text{Ra}^{1/4} \quad (\text{Ra} < 10^7)$$ $$\text{Nu} = 0.15 \text{Ra}^{1/3} \quad (\text{Ra} \ge 10^7)$$ Correlations for lower surface of hot plate / upper surface of cold plate: $$\text{Nu} = 0.27 \text{Ra}^{1/4}$$
Academic References:
- Incropera, F. P., & DeWitt, D. P. (2011). Fundamentals of Heat and Mass Transfer. 7th Edition, John Wiley & Sons.
- Churchill, S. W., & Chu, H. S. (1975). Correlating Equations for Laminar and Turbulent Free Convection on a Vertical Plate. Int. J. Heat Mass Transfer.
Worked Engineering Example
A $0.5\text{ m} \times 0.3\text{ m}$ flat plate tilted at $30^\circ$ from vertical is maintained at $80^\circ\text{C}$ in ambient air at $25^\circ\text{C}$. Find the convection coefficient ($T_f = 52.5^\circ\text{C}$).
Step-by-step Solution:
1. Evaluate air properties at film temperature:
- $\nu = 1.83 \times 10^{-5}\text{ m}^2\text{/s}, \alpha = 2.6 \times 10^{-5}\text{ m}^2\text{/s}, k = 0.028\text{ W/m·K}, Pr = 0.71, \beta = 1/325.65 = 0.00307\text{ K}^{-1}$.
2. Calculate effective gravity:
$$g_{eff} = 9.81 \times \cos(30^\circ) = 8.495\text{ m/s}^2$$ 3. Compute Rayleigh number with $g_{eff}$ ($L_c = 0.5\text{ m}$):
$$\text{Ra}_L = \frac{g_{eff} \beta (T_s - T_a) L_c^3}{\nu \alpha} = \frac{8.495 \times 0.00307 \times (80-25) \times 0.5^3}{1.83 \times 10^{-5} \times 2.6 \times 10^{-5}} = 3.77 \times 10^8$$ 4. Compute Nusselt number using Churchill & Chu correlation:
$$\text{Nu}_L = \left\{0.825 + \frac{0.387 \times (3.77 \times 10^8)^{1/6}}{\left[1 + (0.492/0.71)^{9/16}\right]^{8/27}}\right\}^2 = \left\{0.825 + \frac{10.37}{1.157}\right\}^2 = 95.8$$ 5. Calculate convective coefficient $h$:
$$h = \frac{\text{Nu}_L k}{L_c} = \frac{95.8 \times 0.028}{0.5} = 5.36\text{ W/m}^2\text{K}$$